iCON Education - Some Basic Concepts of Chemistry

2010 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2010 Paper 2 Offline

Silver (atomic weight = 108 g mol^-1) has a density of 10.5 g.cm^-3. The number of silver atoms on a surface of area 10^-12 m² can be expressed in scientific notation as y $\times$ 10^x. The value of x is?

1984 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1984

2.68 $\times$ 10^-3 moles of a solution containing an ion Aⁿ⁺ require 1.61 $\times$ 10^-3 moles of $MnO_4^-$ for the oxidation of An⁺ to $AO_3^-$ in acid medium. What is the value of n?

2022 JEE Advanced MSQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Advanced 2022 Paper 2 Online

To check the principle of multiple proportions, a series of pure binary compounds $\left(\mathrm{P}_{\mathrm{m}} \mathrm{Q}_{\mathrm{n}}\right)$ were analyzed and their composition is tabulated below. The correct option(s) is(are)

Compound	Weight % of $\mathrm{P}$	Weight % of $\mathrm{Q}$
1	50	50
2	44.4	55.6
3	40	60

A.

If empirical formula of compound 3 is $P_{3} Q_{4}$, then the empirical formula of compound 2 is $\mathrm{P}_{3} \mathrm{Q}_{5}$.

B.

If empirical formula of compound 3 is $\mathrm{P}_{3} \mathrm{Q}_{2}$ and atomic weight of element $\mathrm{P}$ is 20 , then the atomic weight of $\mathrm{Q}$ is 45 .

C.

If empirical formula of compound 2 is $P Q$, then the empirical formula of the compound $\mathbf{1}$ is $\mathrm{P}_{5} \mathrm{Q}_{4}$.

D.

If atomic weight of $\mathrm{P}$ and $\mathrm{Q}$ are 70 and 35 , respectively, then the empirical formula of compound $\mathbf{1}$ is $\mathrm{P}_{2} \mathrm{Q}$.

Correct Answer: B,C

Explanation:

(A) If emperical formula of compound 3 is $P_3 Q_4$ then its molar ratio will be $\frac{40}{3} : \frac{60}{4}$

$ =\frac{40}{3} \times \frac{4}{60}=\frac{16}{18}=0.88 $

If empirical formula of compound 2 is $\mathrm{P}_3 \mathrm{Q}_5$, then its molar ratio

$ \begin{aligned} & =\frac{44.4}{2} : {\frac{55.6}{5}} \\\\ & =\frac{44.4}{2} \times \frac{5}{55.6}=2 \end{aligned} $

Since molar ratio of both the compound is not equal

So, option (A) is not correct.

(B) If empirical formula of compound 3 is $\mathrm{P}_3 \mathrm{Q}_2$, i.e.,

$ \begin{aligned} & \frac{40}{M_P}: \frac{60}{M_Q}=\frac{3}{2} \\\\ & \Rightarrow \frac{40}{M_P} \times \frac{M_Q}{60}=\frac{3}{2} \end{aligned} $

$ \begin{aligned} \Rightarrow \frac{4 \mathrm{M}_{\mathrm{Q}}}{6 \mathrm{M}_{\mathrm{P}}} & =\frac{3}{2} \\\\ \Rightarrow \frac{\mathrm{M}_{\mathrm{Q}}}{\mathrm{M}_{\mathrm{P}}} & =\frac{3}{2} \times \frac{6}{4}=\frac{9}{4} \end{aligned} $

and $\mathrm{M}_{\mathrm{P}}=20$ (given)

So, $ \frac{\mathrm{M}_{\mathrm{Q}}}{20}=\frac{9}{4}$

$ \begin{aligned} & \mathrm{M}_{\mathrm{Q}}=\frac{9 \times 26^5}{4} \\\\ & \mathrm{M}_{\mathrm{Q}}=45 \end{aligned} $

So, option (B) is correct.

(C) If empirical formula of compound 2 is PQ, So the molar ratio is

$ \frac{44.4}{1}: \frac{55.6}{1}=\frac{44.4}{55.6}=0.79 \sim 0.8 $

Empirical formula of compound 1 is $P_5 Q_4$

so the molar ratio is $\frac{50}{5} : { } \frac{50}{4}$.

$ =\frac{50}{5} \times \frac{4}{50}=\frac{4}{5}=0.8 $

Since, molar ratio of both the compound is equal hence, state $(\mathrm{C})$ is correct.

So, option (C) is correct.

(D) $\mathrm{M}_{\mathrm{P}}=70, \mathrm{M}_{\mathrm{Q}}=35$

Molar ratio of compound 1 is

$ \begin{aligned} \frac{50}{M_{\mathrm{P}}}: \frac{50}{\mathrm{M}_{\mathrm{Q}}} & =\frac{50}{70}: \frac{50}{35} \\\\ & =\frac{50}{70} \times \frac{35}{50}=\frac{1}{2} \end{aligned} $

Hence, empirical formula of compound $\mathrm{PQ}_2$.

So, option (D) is incorrect.

2011 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2011 Paper 1 Offline

Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the solution is

A.

1.78 M

B.

2.00 M

C.

2.05 M

D.

2.22 M

Correct Answer: C

Explanation:

To find the molarity of the urea solution, we will follow these steps:

1. Calculate the number of moles of urea using its molecular weight.

2. Determine the volume of the solution using its mass and density.

3. Use the number of moles and volume of the solution to calculate the molarity.

Step 1: Calculate the number of moles of urea.

The number of moles of a substance $ n $ is determined by dividing the mass of the substance ($ m $) by its molecular weight ($ M $):

$ n = \frac{m}{M} $

For urea, $ m = 120 \text{ g} $ and $ M = 60 \text{ g/mol} $, thus:

$ n_{urea} = \frac{120 \text{ g}}{60 \text{ g/mol}} $

$ n_{urea} = 2 \text{ moles} $

Step 2: Determine the volume of the solution.

The volume $ V $ of the solution can be found by dividing the mass of the solution by its density $ \rho $:

$ V = \frac{mass_{solution}}{\rho} $

The mass of the solution is the sum of the mass of urea and the mass of water. Thus:

$ mass_{solution} = mass_{urea} + mass_{water} = 120 \text{ g} + 1000 \text{ g} $

$ mass_{solution} = 1120 \text{ g} $

The density of the solution $ \rho $ is given as 1.15 g/mL. Therefore, the volume in milliliters (which is equivalent to cubic centimeters) is:

$ V = \frac{1120 \text{ g}}{1.15 \text{ g/mL}} $

$ V = 973.91 \text{ mL} $

To convert milliliters to liters (since molarity is defined in terms of liters), we divide by 1000:

$ V = 0.97391 \text{ L} $

Step 3: Calculate the molarity.

Molarity ($ M $) is defined as the number of moles of solute divided by the volume of solution in liters:

$ M = \frac{n_{solute}}{V_{solution}} $

$ M = \frac{2 \text{ moles}}{0.97391 \text{ L}} $

$ M \approx 2.05 \text{ M} $

Therefore, the molarity of the urea solution is approximately 2.05 M, which corresponds to Option C.

2009 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2009 Paper 1 Offline

Given that the abundances of isotopes ⁵⁴Fe, ⁵⁶Fe and ⁵⁷Fe are 5%, 90% and 5%, respectively, the atomic mass of Fe is :

A.

55.85

B.

55.95

C.

55.75

D.

56.05

2007 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2007 Paper 2 Offline

Consider a titration of potassium dichromate solution with acidified Mohr’s salt solution using diphenylamine as indicator. The number of moles of Mohr’s salt required per mole of dichromate is

A.

3

B.

4

C.

5

D.

6

2003 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2003 Screening

Mixture X = 0.02 mol of [Co(NH₃)₅SO₄]Br and 0.02 mol of [Co(NH₃)₅Br]SO₄ was prepared in 2 litre of solution.
1 litre of mixture X + excess AgNO₃ $ \to $ Y.
1 litre of mixture X + excess BaCl₂ $ \to $ Z
No. of moles of Y and Z are

A.

0.01, 0.01

B.

0.02, 0.01

C.

0.01, 0.02

D.

0.02, 0.02

2003 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2003 Screening

Which has maximum number of atoms?

A.

24 g of C(12)

B.

56g of Fe(56)

C.

27g of Al(27)

D.

108g of Ag(108)

Correct Answer: A

Explanation:

To determine which option has the maximum number of atoms, we will use Avogadro's number and the concept of moles. Avogadro's number ($6.02 \times 10^{23}$ atoms/mol) represents the number of atoms in one mole of any substance. The number of moles of a substance is calculated by dividing the given mass of the substance by its molar mass (which is given in grams per mole and is numerically equal to the element's atomic mass for pure elements).

Let's calculate the number of moles for each option:

Option A: 24 g of C (Carbon, atomic mass = 12 g/mol)

The number of moles of C = $\frac{24 \text{ g}}{12 \text{ g/mol}} = 2 \text{ moles}$

Therefore, the number of atoms in 24 g of C = $2 \text{ moles} \times 6.02 \times 10^{23} \text{ atoms/mol} = 1.204 \times 10^{24}$ atoms

Option B: 56 g of Fe (Iron, atomic mass = 56 g/mol)

The number of moles of Fe = $\frac{56 \text{ g}}{56 \text{ g/mol}} = 1 \text{ mole}$

Therefore, the number of atoms in 56 g of Fe = $1 \text{ mole} \times 6.02 \times 10^{23} \text{ atoms/mol} = 6.02 \times 10^{23}$ atoms

Option C: 27 g of Al (Aluminum, atomic mass = 27 g/mol)

The number of moles of Al = $\frac{27 \text{ g}}{27 \text{ g/mol}} = 1 \text{ mole}$

Therefore, the number of atoms in 27 g of Al = $1 \text{ mole} \times 6.02 \times 10^{23} \text{ atoms/mol} = 6.02 \times 10^{23}$ atoms

Option D: 108 g of Ag (Silver, atomic mass = 108 g/mol)

The number of moles of Ag = $\frac{108 \text{ g}}{108 \text{ g/mol}} = 1 \text{ mole}$

Therefore, the number of atoms in 108 g of Ag = $1 \text{ mole} \times 6.02 \times 10^{23} \text{ atoms/mol} = 6.02 \times 10^{23}$ atoms

Comparing the number of atoms in each option, Option A with 1.204 $\times 10^{24}$ atoms has the maximum number of atoms.

2002 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2002 Screening

How many moles of electron weigh one kilogram?

A.

$6.023 \times {10^{23}}$

B.

${1 \over {9.108}} \times {10^{31}}$

C.

${{6.023} \over {9.108}} \times {10^{54}}$

D.

${1 \over {9.108 \times 6.023}} \times {10^8}$

Correct Answer: D

Explanation:

To find out how many moles of electrons weigh one kilogram, we need to understand the concept of the mole and the mass of an electron.

A mole is defined as the amount of substance that contains as many entities (atoms, molecules, ions, electrons, etc.) as there are atoms in 12 grams of carbon-12. This number is known as Avogadro's number, $N_A = 6.023 \times 10^{23}$ entities per mole.

The mass of an electron is approximately $9.109 \times 10^{-31}$ kg per electron.

The number of moles ($n$) is given by the equation:

$n = \frac{m}{M}$

where $m$ is the mass of the substance and $M$ is the molar mass of the substance.

In the case of electrons, the mass ($m$) we're considering is 1 kilogram, and we need to find the equivalent mass in terms of moles of electrons.

However, instead of using the molar mass as we would for atoms or molecules, we directly relate the mass of the electrons to Avogadro's number since we're dealing with elementary particles.

The calculation involves finding out how many electrons make up 1 kg, and then calculating how many moles of electrons that represents.

First, find the number of electrons in 1 kg:

$\text{Number of electrons} = \frac{1 \, \text{kg}}{(9.109 \times 10^{-31} \, \text{kg/electron})} = \frac{1}{9.109 \times 10^{-31}}$

Then, to find out how many moles this number of electrons represents, we divide by Avogadro's number:

$\text{Moles of electrons} = \frac{1}{9.109 \times 10^{-31}} \times \frac{1}{6.023 \times 10^{23}}$

Simplifying this gives us:

$\text{Moles of electrons} = \frac{1}{9.109} \times 10^{31} \times \frac{1}{6.023 \times 10^{23}} = \frac{1}{9.109 \times 6.023} \times 10^{8}$

Hence, the option that correctly represents the number of moles of electrons that weigh one kilogram is:

Option D: ${1 \over {9.108 \times 6.023}} \times {10^8}$

2001 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2001 Screening

In the standardization of Na₂S₂O₃ using K₂Cr₂O₇ by iodometry, the equivalent weight of K₂Cr₂O₇ is:

A.

(molecular weight)/2

B.

(molecular weight)/6

C.

(molecular weight)/3

D.

Same as molecular weight

2001 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2001 Screening

An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250ml. the volume of 0.1N NaOH required to completely neutralize 10ml of this solution is :

A.

40ml

B.

20ml

C.

10ml

D.

4ml

Correct Answer: A

Explanation:

To solve this problem, we need to use the concept of normality and the stoichiometry of the reaction between oxalic acid and sodium hydroxide. Oxalic acid dihydrate (C2H2O4.2H2O) reacts with sodium hydroxide (NaOH) in a 1:2 molar ratio because oxalic acid is a diprotic acid, meaning each molecule of oxalic acid can donate two protons.

First, we need to calculate the number of moles of oxalic acid in the solution:

The molar mass of oxalic acid dihydrate (C2H2O4.2H2O) is calculated as follows:

C: 12.01 g/mol * 2 = 24.02 g/mol

H: 1.01 g/mol * 4 = 4.04 g/mol

O: 16.00 g/mol * 6 = 96.00 g/mol

Total: 126.06 g/mol

Given mass of oxalic acid dihydrate = 6.3 g

Moles of oxalic acid dihydrate = $\frac{\text{mass}}{\text{molar mass}} = \frac{6.3 g}{126.06 g/mol} = 0.05 \text{ mol}$

Since the solution is made up to 250 mL, the molarity (M) of this solution is calculated as follows:

$M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} = \frac{0.05 \text{ mol}}{0.25 \text{ L}} = 0.2 \text{ M}$

When 10 mL of this solution reacts with NaOH, the volume of NaOH required to completely neutralize the solution can be found using the stoichiometry of the reaction:

The reaction between oxalic acid (H2C2O4) and NaOH is represented by the balanced chemical equation:

$H_2C_2O_4 + 2NaOH \rightarrow Na_2C_2O_4 + 2H_2O$

This indicates that 1 mole of oxalic acid reacts with 2 moles of NaOH. Therefore, the moles of oxalic acid in 10 mL of the solution can be found as:

$\text{Moles of oxalic acid} = 0.2 M * 0.01 L = 0.002 \text{ mol}$

Since each mole of oxalic acid requires 2 moles of NaOH for complete neutralization:

$\text{Moles of NaOH required} = 0.002 \text{ mol of oxalic acid} * 2 = 0.004 \text{ mol of NaOH}$

The normality (N) of NaOH solution is given as 0.1N. Because NaOH is monovalent, its normality is equal to its molarity in this context. Hence, the volume of NaOH solution required can be calculated using the formula:

$V = \frac{\text{moles of NaOH}}{\text{Normality of NaOH solution}} = \frac{0.004 \text{ mol}}{0.1 N} = 0.04 L = 40 mL$

Therefore, the correct answer is Option A: 40mL.

1991 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1991

Read the following statement and explanation and answer as per the options given below:

STATEMENT (S): In the titration of Na₂CO₃ with HCl using methyl orange indicator, the volume required at the equivalence point is twice that of the acid required using phenolphthalein indicator.

EXPLANATION (E): Two moles of HCl are required for the complete neutralization of one mole of Na₂CO₃

A.

Both S and E are true, and E is the correct explanation of S.

B.

Both S and E are true, and E is not the correct explanation of S.

C.

S is true but E is false

D.

S is false but E is true

Correct Answer: B

Explanation:

The correct answer is Option B : Both S and E are true, and E is not the correct explanation of S.

Let us examine both the statement and the explanation :

STATEMENT (S) : In the titration of Na₂CO₃ with HCl using methyl orange indicator, the volume required at the equivalence point is twice that of the acid required using phenolphthalein indicator.

This statement is true. When titrating Na₂CO₃ (sodium carbonate), which is a diprotic base, there are two stages of the neutralization reaction. The first stage is the neutralization of carbonate ion (CO₃^2-) to bicarbonate ion (HCO₃^-), which is indicated by phenolphthalein changing color. Phenolphthalein changes color at a pH around 8.3, which corresponds to the first endpoint of the titration:

$ \text{Na}_2\text{CO}_3 + \text{HCl} \rightarrow \text{NaHCO}_3 + \text{NaCl} $

The second stage is the neutralization of bicarbonate ion to carbonic acid (H₂CO₃), which then decomposes into water and carbon dioxide. Methyl orange, which changes color at a pH of about 3.1 to 4.4, is used to indicate this second endpoint :

$ \text{NaHCO}_3 + \text{HCl} \rightarrow \text{H}_2\text{CO}_3 + \text{NaCl} $

$ \text{H}_2\text{CO}_3 \rightarrow \text{H}_2\text{O} + \text{CO}_2 $

Therefore, when using methyl orange, the total volume of HCl required for complete neutralization (both stages) is indeed twice the amount needed to reach the first endpoint with phenolphthalein.

EXPLANATION (E) : Two moles of HCl are required for the complete neutralization of one mole of Na₂CO₃.

This explanation is true. As evident from the chemical equations above, one mole of Na₂CO₃ requires one mole of HCl to convert to NaHCO₃, and then another mole of HCl to neutralize the NaHCO₃ further into H₂CO₃ followed by decomposition to water and CO₂. Therefore, the total is two moles of HCl per mole of Na₂CO₃.

Although both the statement and the explanation are true, the explanation does not adequately explain why the volume of HCl is doubled with methyl orange as the indicator compared to phenolphthalein. The correct explanation should involve the different end points detected by the two indicators (phenolphthalein indicating the first stage, methyl orange the second stage), rather than the overall stoichiometry of the reaction.

1991 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1991

The oxidation state of the most electronegative element in the products of the reaction. BaO₂ with dil. H₂SO₄ are

A.

0 and -1

B.

-1 and -2

C.

-2 and 0

D.

-2 and -1

1988 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1988

In which mode of expression, the concentration of a solution remains independent of temperature?

A.

Molarity

B.

Normality

C.

Formality

D.

Molality

1986 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1986

A molal solution is one that contains one mole of solute in

A.

1000 g of solvent

B.

1.0 L of solvent (1986, 1M)

C.

1.0 L of solution

D.

22.4 L of solution

1981 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1981

If 0.50 mol of BaCl₂ is mixed with 0.20 mol of Na₃PO₄, the maximum amount of Ba₃(PO₄)₂ that can be formed is

A.

0.70

B.

0.50

C.

0.20

D.

0.10

Correct Answer: D

Explanation:

To determine the maximum amount of $ \text{Ba}_3(\text{PO}_4)_2 $ that can be formed when $0.50 \, \text{mol}$ of $ \text{BaCl}_2 $ is mixed with $0.20 \, \text{mol}$ of $ \text{Na}_3\text{PO}_4 $, we need to look at the stoichiometry of the reaction between barium chloride and sodium phosphate. The balanced chemical equation for the formation of barium phosphate and sodium chloride is:

$ \text{3BaCl}_2 + 2\text{Na}_3\text{PO}_4 \rightarrow \text{Ba}_3(\text{PO}_4)_2 + 6\text{NaCl} $

This equation tells us that 3 moles of barium chloride react with 2 moles of sodium phosphate to produce 1 mole of barium phosphate. From this, we can calculate the mole ratio needed for the complete reaction:

Ratio for $ \text{BaCl}_2 : \text{Ba}_3(\text{PO}_4)_2 = 3:1 $

Ratio for $ \text{Na}_3\text{PO}_4 : \text{Ba}_3(\text{PO}_4)_2 = 2:1 $

Given amounts are $0.50 \, \text{mol}$ of $ \text{BaCl}_2 $ and $0.20 \, \text{mol}$ of $ \text{Na}_3\text{PO}_4 $. To find out which reactant is the limiting reagent, we calculate how much product each could produce if it were completely consumed:

For $ \text{BaCl}_2 $:

$ \frac{\text{Moles of BaCl}_2}{\text{Mole ratio for BaCl}_2} = \frac{0.50 \, \text{mol}}{3} = 0.167 \, \text{mols of } \text{Ba}_3(\text{PO}_4)_2 $

For $ \text{Na}_3\text{PO}_4 $:

$ \frac{\text{Moles of Na}_3\text{PO}_4}{\text{Mole ratio for Na}_3\text{PO}_4} = \frac{0.20 \, \text{mol}}{2} = 0.1 \, \text{mols of } \text{Ba}_3(\text{PO}_4)_2 $

From the calculations, $ \text{Na}_3\text{PO}_4 $ yields less $ \text{Ba}_3(\text{PO}_4)_2 $ than $ \text{BaCl}_2 $ does; therefore, $ \text{Na}_3\text{PO}_4 $ is the limiting reactant. This means the maximum amount of $ \text{Ba}_3(\text{PO}_4)_2 $ that can be formed from the given amounts of reactants is $0.10 \, \text{mol}$.

So, the answer is Option D: 0.10.

1979 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1979

2.76 g of silver carbonate on being strongly heated yields a residue weighing

A.

2.16 g

B.

2.48 g

C.

2.32 g

D.

2.64 g

1979 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1979

The total number of electrons is one molecule of carbon dioxide is

A.

22

B.

44

C.

66

D.

88

1979 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1979

The largest number of molecules is in

A.

32 g of water

B.

28 g of carbon monoxide

C.

46 g of ethyl alcohol

D.

54 g of nitrogen pentoxide

1979 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1979

A gaseous mixture contains oxygen and nitrogen in the ratio of 1 : 4 by weight. Therefore the ratio of their no of molecules is

A.

1 : 4

B.

1 : 8

C.

7 : 32

D.

3 : 16

1978 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1978

A compound was found to contain nitrogen and oxygen in the ratio 28 gm and 80 gm respectively. The formula of compound is

A.

NO

B.

N₂O₃

C.

N₂O₅

D.

N₂O₄

1978 JEE Advanced MCQ

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1978

27 g of Al will react completely with how many grams if oxygen?

A.

8 g

B.

16 g

C.

32 g

D.

24 g

Correct Answer: D

Explanation:

To determine how many grams of oxygen will react completely with 27 grams of aluminum (Al), we first need to look at the balanced chemical equation for the reaction between aluminum and oxygen to form aluminum oxide ($Al_2O_3$). The balanced equation is:

$4Al + 3O_2 \rightarrow 2Al_2O_3$

This equation tells us that 4 moles of aluminum react with 3 moles of oxygen gas to produce 2 moles of aluminum oxide.

First, we need to determine the molar mass of aluminum (Al) and oxygen (O₂) to proceed with the calculations.

The molar mass of aluminum (Al) is approximately $27 \, g/mol$.

The molar mass of oxygen gas (O₂) is $32 \, g/mol$ (since the molar mass of an oxygen atom is approximately 16 g/mol, and $O_2$ has two oxygen atoms).

Given that we have 27 g of aluminum:

The number of moles of aluminum ($n_{Al}$) can be calculated as follows:

$n_{Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} = \frac{27 \, g}{27 \, g/mol} = 1 \, mol$

According to the balanced equation, 4 moles of Al react with 3 moles of O₂. Therefore, to find out how many moles of O₂ react with 1 mole of Al, we set a proportion:

$\frac{4 \, mol \, Al}{3 \, mol \, O_2} = \frac{1 \, mol \, Al}{x \, mol \, O_2}$

Solving for $x$, we find:

$x = \frac{3 \, mol \, O_2}{4 \, mol \, Al} = \frac{3}{4} \, mol \, O_2$

To find out the mass of oxygen that will react, we use the molar mass of oxygen:
$mass \, O_2 = n_{O_2} \times \text{molar mass of }\, O_2 = \frac{3}{4} \, mol \times 32 \, g/mol = 24 \, g$

Therefore, 27 g of aluminum will react completely with 24 g of oxygen. The correct answer is Option D.

2003 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2003

Calculate the molarity of water if it's density is 1000 kg/m³

2001 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 2001

Hydrogen peroxide solution (20 ml) reacts quantitatively with a solution of KMnO₄ solution is just decolourised by 10 ml of MnSO₄ in neutral medium simultaneously forming a dark brown precipitate of hydrated MnO₂. The brown precipitated is dissolved in 10 ml of 0.2 M sodium oxalate under boiling condition in the presence of dilute H₂SO₄. Write the balanced equations involved in the reactions and calculate the molarity of H₂O₂.

Correct Answer: Molarity : 0.1

Explanation:

Let the molarity of $\mathrm{KMnO}_4$ be $M_1$. Moles of $\mathrm{KMnO}_4$ in $20 \mathrm{ml}$

$ =\frac{20 \times M_1}{1000} $

$n$-factor of $\mathrm{KMnO}_4$ when it reacts with $\mathrm{MnSO}_4$ is 3.

$\therefore$ Equivalent of $\mathrm{KMnO}_4$ reacting with $\mathrm{MnSO}_4=\frac{20 \times M_1}{1000} \times 3$

$\therefore$ Equivalent of $\mathrm{MnSO}_4$ reacting with $\mathrm{KMnO}_4=\frac{20 \times M_1}{1000} \times 3$

Since $\mathrm{MnSO}_4$ has $n$-factor 2, the mole of $\mathrm{MnSO}_4$ reacting

$ =\frac{20 \times M_1}{1000} \times \frac{3}{2} $

Total mole of $\mathrm{MnO}_2$ produced $=$ mole of $\mathrm{KMnO}_4+$ mole of $ \mathrm{MnSO}_4 $

Equivalent of $\mathrm{Na}_2 \mathrm{C}_2 \mathrm{O}_4$ reacting with $\mathrm{MnO}_2=\frac{10 \times 0.2}{1000} \times 2$

$\therefore$ Equivalent of $\mathrm{MnO}_2=\frac{10 \times 0.2}{1000} \times 2$

Mole of $\mathrm{MnO}_2$ reacting with sodium oxalate $=\frac{10 \times 0.2 \times 2}{1000 \times 2}$

[as $n$-factor for $\mathrm{MnO}_2$ is 2].

Therefore, $\frac{20 \times M_1}{1000}+\left[\frac{20 \times M_1}{1000} \times \frac{3}{2}\right]=\frac{10 \times 0.2}{1000} ; M_1=0.04$

Equivalent of $\mathrm{KMnO}_4$ reacting with $\mathrm{H}_2 \mathrm{O}_2=\frac{20 \times 0.04}{1000} \times 5$ $ =0.004 $

If molarity of $\mathrm{H}_2 \mathrm{O}_2$ is $M_2$, then $=\frac{20 \times M_2 \times 2}{1000}=0.004$

$\therefore M_2=0.1 \mathrm{M}$

The reactions involved are:

$2 \mathrm{KMnO}_4+5 \mathrm{H}_2\mathrm{O}_2+3 \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{K}_2\mathrm{SO}_4+2 \mathrm{MnSO}_4+8 \mathrm{H}_2\mathrm{O}+5 \mathrm{O}_2$,

$2 \mathrm{KMnO}_4+3 \mathrm{MnSO}_4+2 \mathrm{H}_2\mathrm{O} \rightarrow 5 \mathrm{MnO}_2+2 \mathrm{H}_2\mathrm{SO}_4+\mathrm{K}_2\mathrm{SO}_4$,

$\mathrm{MnO}_2+\mathrm{Na}_2\mathrm{C}_2\mathrm{O}_4+2 \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{MnSO}_4+2 \mathrm{CO}_2+\mathrm{Na}_2\mathrm{SO}_4+2 \mathrm{H}_2\mathrm{O}$.

1999 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1999

How many millilitres of 0.5 M H₂SO₄ are needed to dissolve 0.5 g of copper(II) carbonate?

Correct Answer: 8.092 ml

Explanation:

To determine how many millilitres of 0.5 M $ \text{H}_2\text{SO}_4 $ are needed to dissolve 0.5 g of copper(II) carbonate $ (\text{CuCO}_3) $, we first need to write the chemical reaction between the copper(II) carbonate and sulfuric acid ( $ \text{H}_2\text{SO}_4 $ ):

$ \text{CuCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{H}_2\text{O} + \text{CO}_2 $

From this reaction, it's observable that 1 mole of copper(II) carbonate reacts with 1 mole of sulfuric acid to produce 1 mole of copper(II) sulfate, water, and carbon dioxide.

Let's calculate the number of moles of copper(II) carbonate in 0.5 g:

The molar mass of $ \text{CuCO}_3 $ is calculated as:

$ \text{Molar mass of } \text{CuCO}_3 = (\text{Molar mass of Cu}) + (\text{Molar mass of C}) + 3 \times (\text{Molar mass of O}) $

$ \text{Molar mass of } \text{CuCO}_3 = (63.55 \text{ g/mol}) + (12.01 \text{ g/mol}) + 3 \times (16.00\text{ g/mol}) $

$ \text{Molar mass of } \text{CuCO}_3 = 63.55 + 12.01 + 48.00 $

$ \text{Molar mass of } \text{CuCO}_3 = 123.56 \text{ g/mol} $

Now, we'll find out the number of moles in 0.5 g of $ \text{CuCO}_3 $:

$ \text{Number of moles of } \text{CuCO}_3 = \frac{\text{mass of } \text{CuCO}_3}{\text{Molar mass of } \text{CuCO}_3} $

$ \text{Number of moles of } \text{CuCO}_3 = \frac{0.5 \text{ g}}{123.56 \text{ g/mol}} $

$ \text{Number of moles of } \text{CuCO}_3 \approx 0.004046 \text{ mol} $

Since the reaction is a 1:1 molar ratio, the moles of sulfuric acid needed would also be $ 0.004046 \text{ mol} $.

The concentration (C) of a solution is given by the formula:

$ C = \frac{n}{V} $

where:

$ C $ is the concentration of the solution in moles per liter (M).

$ n $ is the number of moles of solute.

$ V $ is the volume of solution in liters.

Rearranging the formula to solve for volume $ V $:

$ V = \frac{n}{C} $

The given concentration of $ \text{H}_2\text{SO}_4 $ is 0.5 M, which means 0.5 moles of $ \text{H}_2\text{SO}_4 $ per liter. Now, let's calculate the volume required for the $ 0.004046 \text{ mol} $ of $ \text{H}_2\text{SO}_4 $:

$ V = \frac{0.004046 \text{ mol}}{0.5 \text{ M}} = \frac{0.004046 \text{ mol}}{0.5 \text{ mol/L}} $

$ V = 0.008092 \text{ L} $

To convert liters to millilitres:

$ V(\text{mL}) = 0.008092 \text{ L} \times 1000 \frac{\text{mL}}{\text{L}} $

$ V(\text{mL}) = 8.092 \text{ mL} $

Therefore, you will need approximately 8.092 mL of 0.5 M $ \text{H}_2\text{SO}_4 $ to completely dissolve 0.5 g of copper(II) carbonate under the assumption that the reaction goes to completion with no side reactions.

1999 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1999

A plant virus is found to consist of uniform cylindrical particles of 150 Å in diameter and 5000 Å long. The specific volume of the virus is 0.75 cm³/g. If the virus is considered to be a single particle, find its molar mass.

1998 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1998

An aqueous solution containing 0.10 g KIO₃ (formula weight = 214.0) was treated with an excess of KI solution. The solution was acidified with HCl. The liberated I₂ consumed 45.0 mL of thiosulphate solution to decolourise the blue strach-iodine complex. Calculate the molarity of the sodium thiosulphate solution.

Correct Answer: 0.0623

Explanation:

To calculate the molarity of the sodium thiosulphate solution, we'll need to apply stoichiometry principles. First, we must determine the number of moles of $ \text{KIO}_3 $ as it will relate to the number of moles of $ \text{I}_2 $ liberated during the reaction. Then we will use the volume of the sodium thiosulphate solution to find its molarity.

The reaction for the iodometry where $ \text{I}_2 $ is liberated from $ \text{KIO}_3 $ and reacted with the excess of $ \text{KI} $ is:

$ \text{KIO}_3 + 5 \text{KI} + 6 \text{HCl} \rightarrow 3 \text{I}_2 + 6 \text{KCl} + 3 \text{H}_2\text{O} $

From the balanced equation, we see that 1 mole of $ \text{KIO}_3 $ produces 3 moles of $ \text{I}_2 $.

Now let's find the number of moles of $ \text{KIO}_3 $:

$ \text{Moles of } \text{KIO}_3 = \frac{\text{mass of } \text{KIO}_3}{\text{molar mass of } \text{KIO}_3} $

$ = \frac{0.10 \text{ g}}{214.0 \text{ g/mol}} $

To get the number of moles, we need to perform the division:

$ = \frac{0.10}{214.0} $

$ = 4.6728972 \times 10^{-4} \text{ moles of } \text{KIO}_3 $

Since 1 mole of $ \text{KIO}_3 $ yields 3 moles of $ \text{I}_2 $, we multiply the moles of $ \text{KIO}_3 $ by 3:

$ \text{Moles of } \text{I}_2 = 3 \times 4.6728972 \times 10^{-4} $

$ = 1.4018692 \times 10^{-3} \text{ moles of } \text{I}_2 $

Next, $ \text{I}_2 $ reacts with sodium thiosulphate ($ \text{Na}_2\text{S}_2\text{O}_3 $) in the following stoichiometric reaction:

$ \text{I}_2 + 2 \text{Na}_2\text{S}_2\text{O}_3 \rightarrow 2 \text{NaI} + \text{Na}_2\text{S}_4\text{O}_6 $

From the reaction, 1 mole of $ \text{I}_2 $ reacts with 2 moles of sodium thiosulphate. Therefore, the moles of sodium thiosulphate required to react with the $ \text{I}_2 $ produced can be calculated as follows:

$ \text{Moles of } \text{Na}_2\text{S}_2\text{O}_3 = 2 \times \text{Moles of } \text{I}_2 $

$ = 2 \times 1.4018692 \times 10^{-3} $

$ = 2.8037384 \times 10^{-3} \text{ moles of } \text{Na}_2\text{S}_2\text{O}_3 $

Now we have to calculate the molarity of the sodium thiosulphate solution. Molarity (M) is defined as the number of moles of solute divided by the volume of solution in liters:

$ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} $

We have the number of moles of sodium thiosulphate and the volume is given as 45.0 mL which needs to be converted to liters:

$ \text{Volume in liters} = 45.0 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.045 \text{ L} $

Substitute the number of moles and the volume into the molarity formula:

$ M = \frac{2.8037384 \times 10^{-3} \text{ moles}}{0.045 \text{ L}} $

Performing the division gives us the molarity:

$ M = \frac{2.8037384 \times 10^{-3}}{0.045} $

$ = 0.0623053 \text{ M} $

So, the molarity of the sodium thiosulphate solution is approximately $ 0.0623 $ M.

1996 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1996

A 3.00 g sample containing Fe₃O₄, Fe₂O₃ and an inert impure substance is treated with excess of KI solution in presence of dilute H₂SO₄. The entire iron is converted into Fe²⁺ along with the liberation of iodine. The resulting solution is diluted to 100 ml, A 20 ml of the diluted solution requires 11.0 ml of 0.5 M Na₂S₂O₃ solution to reduce the iodine present. A 50 ml of the diluted solution, after complete extraction of the iodine requires 12.80 ml of 0.25 M KMnO₄ solution in the dilute H₂SO₄ medium for the oxidation of Fe²⁺. Calculate the percentages of Fe₂O₃ and Fe₃O₄ in the original sample.

Correct Answer: 49.33%, 34.8%

Explanation:

Let the original sample contains $x$ millimol of $\mathrm{Fe}_3 \mathrm{O}_4$ and $y$ millimol of $\mathrm{Fe}_2 \mathrm{O}_3$. In the first phase of reaction,

$ \begin{aligned} & \mathrm{Fe}_3 \mathrm{O}_4+\mathrm{I}^{-} \longrightarrow 3 \mathrm{Fe}^{2+}+\mathrm{I}_2\left(n \text {-factor of } \mathrm{Fe}_3 \mathrm{O}_4=2\right) \\\\ & \mathrm{Fe}_2 \mathrm{O}_3+\mathrm{I}^{-} \longrightarrow 2 \mathrm{Fe}^{2+}+\mathrm{I}_2\left(n \text {-factor of } \mathrm{Fe}_2 \mathrm{O}_3=2\right) \end{aligned} $

$\begin{aligned} \Rightarrow \text { Meq of } \mathrm{I}_2 \text { formed } & =\mathrm{Meq}\left(\mathrm{Fe}_3 \mathrm{O}_4+\mathrm{Fe}_2 \mathrm{O}_3\right) =\text { Meq of hypo required } \\\\ \Rightarrow 2 x+2 y & =11 \times 0.5 \times 5=27.5 ...........(i)\end{aligned}$

Now, total millimol of $\mathrm{Fe}^{2+}$ formed $=3 x+2 y$. In the reaction

$ \begin{array}{ll} \mathrm{Fe}^{2+}+\mathrm{MnO}_4^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{Mn}^{2+} \\\\ n \text {-factor of } \mathrm{Fe}^{2+}=1 \\\\ \Rightarrow \text { Meq of } \mathrm{MnO}_4^{-}=\text {Meq of } \mathrm{Fe}^{2+} \\\\ \Rightarrow 3 x+2 y=12.8 \times 0.25 \times 5 \times 2=32 ...........(ii) \end{array} $

Solving Eqs. (i) and (ii), we get

$ \begin{aligned} x=4.5 \text { and } y=9.25 \end{aligned} $

$\begin{aligned} \Rightarrow \text { Mass of } \mathrm{Fe}_3 \mathrm{O}_4 & =\frac{4.5}{1000} \times 232=1.044 \mathrm{~g} \\\\ \% \text { mass of } \mathrm{Fe}_3 \mathrm{O}_4 & =\frac{1.044}{3} \times 100=34.80 \% \\\\ \text { Mass of } \mathrm{Fe}_2 \mathrm{O}_3 & =\frac{9.25}{1000} \times 160=1.48 \mathrm{~g} \\\\ \% \text { mass of } \mathrm{Fe}_2 \mathrm{O}_3 & =\frac{1.48}{3} \times 100=49.33 \%\end{aligned}$

1994 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1994

8.0575 $\times$ 10^-2 kg of Glauber's salt is dissolved in water to obtain 1 dm³ of a solution of density 1077.2 kg.m^-3. Calculate the molarity, molality and mole fraction Na₂SO₄ in the solution.

Correct Answer: 0.25 M, 0.24 m, 4.3 $$\times$$ 10-3

Explanation:

Glauber's salt is $\mathrm{Na}_2 \mathrm{SO}_4 \cdot 10 \mathrm{H}_2 \mathrm{O}$.

Molecular weight of $\mathrm{Na}_2 \mathrm{SO}_4=142$

Molecular weight of $\mathrm{Na}_2 \mathrm{SO}_4 \cdot 10 \mathrm{H}_2 \mathrm{O}=322$

Weight of Glauber's salt taken $=8.0575 \times 10^{-2} \mathrm{~kg}=80.575 \mathrm{~g}$

Weight of anhydrous $\mathrm{Na}_2 \mathrm{SO}_4$ in $322 \mathrm{~g}$ of Glauber's salt

$=142 / 322 \mathrm{~g}$

$\therefore$ Weight of anhydrous $\mathrm{Na}_2 \mathrm{SO}_4$ in $80.575 \mathrm{~g}$ of Glauber's salt

$ =\frac{142}{322} \times 80.575=35.53 \mathrm{~g} $

Hence number of moles of $\mathrm{Na}_2 \mathrm{SO}_4$ per $\mathrm{dm}^3$ of solution

$ =\frac{35.53}{142}=0.25 $

So, the molarity of solution is $0.25 \mathrm{M}$

$ \begin{aligned} \text { Density of solution } & =1077.2 \mathrm{~kg} \mathrm{~m}^{-3} \\\\ & =\frac{1077.2 \times 10^3}{10^6} \mathrm{~g} \mathrm{~cm}^{-3}=1.0772 \mathrm{~g} \mathrm{~cm}^{-3} \end{aligned} $

$\begin{aligned} & \text { Weight of solution }=\text { volume } \times \text { density } \\\\ & =1000 \times 1.0772 \mathrm{~g}=1077.2 \mathrm{~g} \\\\ & \text { Weight of water }=(1077.2-35.53)=1041.67 \mathrm{~g} \\\\ & \text { Molality of solution }=\frac{0.25}{1041.67} \times 1000=0.24 \mathrm{~m}\end{aligned}$

$\begin{aligned} & \text { Number of moles of water in solution }=\frac{1041.67}{18}=57.87 \\\\ & \text { Mole fraction of } \mathrm{Na}_2 \mathrm{SO}_4=\frac{\text { Moles of } \mathrm{Na}_2 \mathrm{SO}_4}{\text { Total number of moles }} \\\\ & \qquad=\frac{0.25}{0.25+57.87}=0.0043 \text { or } 4.3 \times 10^{-3}\end{aligned}$

1994 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1994

The composition of a sample of Wustite is Fe_0.93O_1.00 what percentage of the iron is present in the form of Fe(III)?

Correct Answer: 15.05%

Explanation:

To determine the percentage of iron present as Fe(III) in the sample of wustite (Fe_0.93O_1.00), we need to assume that the rest of the iron that is not Fe(II) is present as Fe(III). This is because wustite, which nominally has the formula FeO, has a non-stoichiometric composition due to the presence of both Fe(II) and Fe(III) ions that maintain charge neutrality in the lattice.

Given the formula Fe_0.93O_1.00, for every one oxygen atom, there are 0.93 irons. In terms of charge, the oxygen anion has a charge of -2. The formula can be interpreted as having 0.93 moles of Fe per mole of wustite, and we need to balance the charges considering the presence of both Fe(II) and Fe(III).

Fe(II) has a charge of +2, and Fe(III) has a charge of +3. If we let x be the fraction of Fe(III), then the fraction of Fe(II) would be 0.93 - x because the total amount of iron is 0.93 moles of iron (Fe). Now, we can set up a charge balance equation:

Charge from Fe(II) + Charge from Fe(III) = Charge from O

(0.93 - x)(+2) + (x)(+3) = 1(+2)

2(0.93 - x) + 3x = 2

1.86 - 2x + 3x = 2

Combining like terms gives:

1.86 + x = 2

x = 2 - 1.86

x = 0.14

So, 0.14 moles of Fe are in the form of Fe(III) per mole of wustite.

To find the percentage of Fe(III), you divide the number of moles of Fe(III) by the total moles of Fe and multiply by 100%:

$ \text{Percentage of Fe(III)} = \left( \frac{x}{0.93} \right) \times 100\% $

$ \text{Percentage of Fe(III)} = \left( \frac{0.14}{0.93} \right) \times 100\% $

$ \text{Percentage of Fe(III)} = (0.1505) \times 100\% $

$ \text{Percentage of Fe(III)} = 15.05\% $

Therefore, 15.05% of the iron is present in the form of Fe(III) in the sample of wustite Fe_0.93O_1.00.

1993 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1993

Upon mixing 45.0 ml. of 0.25 M lead nitrate solution with 25.0 ml. of 0.10M chromic sulphate solution, precipitation of lead sulphate takes place. How many moles of lead sulphate are formed? Also, calculate the molar concentrations of the species left behind in the final solution. Assume that lead sulphate is completely insoluble.

1992 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1992

A 2.0 g sample of a mixture containing sodium carbonate, sodium bicarbonate and sodium suplphate is gently heated till the evolution of CO₂ ceases. The volume of CO₂ at 750 mm Hg pressure and at 298K is measured to be 123.9 ml. A 1.5g of the same sample requires 150 ml. of (M/10) HCl for complete neutralisation. Calculate the % composition of the components of the mixture.

Correct Answer: Na2CO3 - 26.5% Na2SO4 - 31.5%

Explanation:

$\mathrm{CO}_2$ is evolved due to following reaction :

$ 2 \mathrm{NaHCO}_3 \longrightarrow \mathrm{Na}_2 \mathrm{CO}_3+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 $

$ \begin{aligned} \text { Moles of } \mathrm{CO}_2 \text { produced } & =\frac{p V}{R T}=\frac{750}{760} \times \frac{123.9}{1000} \times \frac{1}{0.082 \times 298} \\\\ & =5 \times 10^{-3} \end{aligned} $

$\begin{aligned} & \Rightarrow \text { Moles of } \mathrm{NaHCO}_3 \text { in } 2 \mathrm{~g} \text { sample }=2 \times 5 \times 10^{-3}=0.01 \\\\ & \Rightarrow \text { millimol of } \mathrm{NaHCO}_3 \text { in } 1.5 \mathrm{~g} \text { sample } \\\\ & \qquad=\frac{0.01}{2} \times 1.5 \times 1000=7.5\end{aligned}$

Let the $1.5 \mathrm{~g}$ sample contain $x$ millimol $\mathrm{Na}_2 \mathrm{CO}_3$, then

$ \begin{array}{rlrl} & 2 x+7.5 =\text { millimol of } \mathrm{HCl}=15 \\\\ & \Rightarrow x =3.75 \end{array} $

$\begin{aligned} \Rightarrow \text { Mass of } \mathrm{NaHCO}_3 =\frac{7.5 \times 84}{1000}=0.63 \mathrm{~g} \\\\ \text { Mass of } \mathrm{Na}_2 \mathrm{CO}_3=\frac{3.75 \times 106}{1000}=0.3975 \mathrm{~g}\end{aligned}$

$\begin{aligned} \Rightarrow \% \text { mass of } \mathrm{NaHCO}_3 =\frac{0.63}{1.50} \times 100=42 \% \\\\ \% \text { mass of } \mathrm{Na}_2 \mathrm{CO}_3 =\frac{0.3975}{1.5} \times 100=26.5 \%\end{aligned}$

$\begin{aligned} \% \text { of } \mathrm{Na}_2 \mathrm{SO}_4 \text { in the sample } & =100-(42+26.5) \\\\ & =100-68.5=31.5 \%\end{aligned}$

1992 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1992

One gram of commercial AgNO₃ is dissolved in 50 ml. of water. It is treated with 50 ml. of a KI solution. The silver iodide thus precipitated is filtered off. Excess of KI in the filterate is titrated with (M/10) KIO₃ solution in presence of 6M HCl till all I^- ions are converted into ICl. It requires 50 ml. of (M/10) KIO₃ solution. 20 ml. of the same stock solution of KI requires 30 ml. of (M/10) KIO₃ under similar conditions. Calculate the percentage of AgNO₃ in the sample.
(Reaction : KIO₃ + 2KI + 6HCl $\to$ 3ICl + 3KCl + 3H₂O)

Correct Answer: 85%

Explanation:

First, let's write down the given data for clarity:

Mass of commercial AgNO₃ = 1 g

Volume of the KI solution used = 50 ml

Volume of (M/10) KIO₃ solution used to titrate excess KI = 50 ml

Volume of (M/10) KIO₃ solution used to titrate 20 ml of KI solution = 30 ml

From the reaction given:

$ KIO_3 + 2KI + 6HCl \to 3ICl + 3KCl + 3H_2O $

It can be seen that 1 mole of KIO₃ reacts with 2 moles of KI. The molarity of KIO₃ solution is also given as (M/10), which means 0.1 M.

First, let's calculate the moles of KIO₃ used to titrate the excess KI present after the precipitation of AgI:

$ \text{Moles of KIO}_3 = \text{Volume} \times \text{Molarity} $

$ = 50 \text{ ml} \times 0.1 \text{ M} $

Since 1 liter is 1000 ml, we convert ml to liters:

$ = \frac{50}{1000} \text{ L} \times \text{0.1 M} $

$ = 0.005 \text{ moles} $

Now, 1 mole of KIO₃ reacts with 2 moles of KI; therefore, 0.005 moles of KIO₃ will react with:

$ 0.005 \text{ moles KIO}_3 \times 2 \text{ moles KI} / 1 \text{ mole KIO}_3 $

$ = 0.01 \text{ moles KI} $

So, the excess amount of KI after the precipitation of AgI is 0.01 moles.

Next, we need to find out the amount of KI present in the 20 ml of KI solution that reacts with 30 ml of the KIO₃ solution.

$ \text{Moles of KIO}_3 (for 20 \text{ ml KI solution}) = \text{Volume} \times \text{Molarity} $

$ = 30 \text{ ml} \times 0.1 \text{ M} $

$ = \frac{30}{1000} \text{ L} \times \text{0.1 M} $

$ = 0.003 \text{ moles KIO}_3 $

This will react with twice the amount of KI:

$ 0.003 \text{ moles KIO}_3 \times 2 \text{ moles KI} / 1 \text{ mole KIO}_3 $

$ = 0.006 \text{ moles KI} $

This amount is present in 20 ml of the KI solution. To find the amount of KI in the initial 50 ml used for the reaction with AgNO₃, we set up a proportion, assuming the KI solution is uniform in concentration:

$ \frac{0.006 \text{ moles KI}}{20 \text{ ml}} = \frac{x \text{ moles KI}}{50 \text{ ml}} $

Solving for x:

$ x = \frac{0.006 \text{ moles KI} \times 50 \text{ ml}}{20 \text{ ml}} $

$ x = 0.015 \text{ moles KI} $

From the reaction between AgNO₃ and KI:

$ AgNO_3 + KI \to AgI \downarrow + KNO_3 $

You can see that 1 mole of AgNO₃ reacts with 1 mole of KI. If 0.01 moles of KI remained after the reaction, the amount of KI that reacted with AgNO₃ is:

$ 0.015 \text{ moles KI (total)} - 0.01 \text{ moles KI (excess)} $

$ = 0.005 \text{ moles KI reacted with AgNO}_3 $

Therefore, the moles of AgNO₃ in the 1 gram commercial sample is also 0.005 moles, since the molar ratio of AgNO₃ to KI is 1:1. Now, let's calculate the mass of 0.005 moles of pure AgNO3:

The molar mass of AgNO₃ = Atomic mass of Ag + Atomic mass of N + 3 x Atomic mass of O

= 108 + 14 + 3 x 16 = 170 g/mol

The mass of 0.005 moles of AgNO₃ is:

$ 0.005 \text{ moles} \times 170 \text{ g/mol} $

$ = 0.85 \text{ grams} $

Finally, to find the percentage of AgNO₃ in the sample, we divide the mass of pure AgNO₃ by the mass of the commercial sample and multiply by 100%.

$ \text{Percentage of AgNO}_3 = \frac{0.8493655 \text{ g}}{1 \text{ g}} \times 100\% $

$ = 85\% $

Thus, the sample contains approximately 85 AgNO₃.

1991 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1991

A 1.0 g sample of Fe₂O₃ solid of 55.2% purity is dissolved in acid and reduced by heating the solution with zinc dust. The resultant solution is cooled and made upto 100.0 ml. An aliquot of 25.0 ml of this solution requires 17.0 ml of 0.0167M solution of an oxidant for titration. Calculate the number of electrons taken by the oxidant in the reaction of the above titration.

1991 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1991

A solution of 0.2 g of a compound containing Cu²⁺ and $C_2O_4^{2-}$ ions on titration with 0.02M $KMnO_4$ in presence of $H_2SO_4$ consumes 22.6 ml. of the oxidant. The resultant solution in neutralized with Na₂CO₃, acidified with dil. acetic acid and treated with excess KI. The liberated iodine requires 11.3 ml of 0.05M Na₂S₂O₃ solution for complete reduction.

Find out the molar ratio of Cu²⁺ to $C_2O_4^{2-}$ in the compound.Write down the balanced redox reactions involved in the above titrations.

1990 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1990

A mixture of H₂C₂O₄ (oxalic acid) and NaHC₂O₄ weighing 2.02 g was dissolved in water and solution made upto one litre. Ten millilitres of the solution required 3.0 ml. of 0.1 N sodium hydroxide solution for complete neutralization. In another experiment, 10.0 ml. of the same solution, in hot dilute sulphuric acid medium. require 4.0 ml. of 0.1 N potassium permanganate solution for complete reaction. Calculate the amount of H₂C₂O₄ and NaHC₂O₄ in the mixture.

1990 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1990

Calculate the molality of 1 litre solution of 93% H₂SO₄ (weight/volume). The density of the solution is 1.84 g/ml

1990 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1990

A solid mixture (5.0 g) consisting of lead nitrate and sodium nitrate was headed below 600^oC until the weight of the residue was constant. If the loss in weight 28.0 percent, find the amount of lead nitrate and sodium nitrate in the mixture.

1989 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1989

An equal volume of a reducing agent is titrated separately with 1M KMnO₄ in acid neutral and alkaline media. The volumes of KMnO₄ required are 20 ml. in acid 33.4 ml. neutral and 100 ml in alkaline media. Find out the oxidation state of manganese in each reduction product. Give the balanced equations for all three half reactions. Find out the volume of 1M K₂Cr₂O₇ consumed; if the same volume of the reducing agent is titrated in acid medium.

1988 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1988

A sugar syrup of weight 214.2 g contains 34.2 g of sugar (C₁₂H₂₂O₁₁). Calculate (i) molal concentration and (ii) mole fraction of sugar in the syrup

1988 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1988

A sample of hydrazine sulphate (N₂H₆SO₄) was dissolved in 100 ml. of water, 10 ml of this solution was reacted with excess of ferric chloride solution and warmed to complete the reaction. Ferrous ion formed was estimated and it required 20 ml of M/50 potassium permanganate solution. Estimate the amount of hydrazine sulphate in one litre of the solution.

Reaction:

4Fe³⁺ + N₂H₄ $\to$ N₂ + 4Fe²⁺ + 4H⁺

$MnO_4^-$ + 4Fe²⁺ + 8H⁺ $\to$ Mn²⁺ + 5Fe³⁺ + 4H₂O

1985 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1985

Five ml of 8N nitric acid, 4.8ml of 5N hydrochloric acid and a certain volume of 17M sulphuric acid are mixed together and made upto 2 litre. Thirty ml of this acid mixture neutralise 42.9 ml of sodium carbonate solution containing one gram of Na₂CO₃.10H₂O in 100 ml of water. Calculate the amount in gram of the sulphate ions in solution.

1983 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1983

4.08 g of a mixture of BaO and an unknown carbonate MCO₃ was heated strongly. The residue weighed 3.64 g. This was dissolved in 100 ml of 1 N HCl. The excess acid required 16 ml of 2.5 N NaOH solution for complete neutralization. Identify the metal M. (At wt. H = 1, C = 12, O = 16, Cl = 35.5, Ba = 138)

1983 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1983

The density of a 3 M Sodium thiosulphate solution (Na₂S₂O₃) is 1.25 g per ml. Calculate

(i) the percentage by weight of sodium thiosulphate,

(ii) the mole fraction of sodium thiosulphate and

(iii) the molalities of Na⁺ and $S_2O_3^{-2}$ ions

1982 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1982

Hydroxylamine reduces iron (III) according to the equation:

2NH₂OH + 4Fe³⁺ $\to$ N₂O(g) $ \uparrow $ + H₂O + 4Fe²⁺ + 4H⁺

Iron (II) thus produced is estimated by titration with a standard permanganate solution. The reaction is :

$MnO_4^-$ + 5Fe²⁺ + 8H⁺ $\to$ Mn²⁺ + 5Fe³⁺ + 4H₂O

A 10 ml sample of hydroxylamine solution was Diluted to 1 litre. 50 ml of this diluted solution was boiled with an excess of iron (III) solution. The resulting solution required 12 ml of 0.02 M KMnO₄ solution for complete oxidation of iron (II). Calculate the weight of hydroxylamine in one litre of the original solution. (H = 1, N = 14, O = 16, K =39, Mn = 55, Fe = 56)

1981 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1981

A 1.00 gm sample of H₂O₂ solution containing X percent H₂O₂ by weight requires X ml of a KMnO₄ solution for complete oxidation under acidic conditions. Calculate the normality of the KMnO₄ solution.

1980 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1980

Find
(i) The total number of neutrons and
(ii) The total mass of neutron in 7 mg of ¹⁴C
(Assume that mass of neutron = mass of hydrogen atom)

1980 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1980

(a) 1 litre of a mixture of CO and CO₂ is taken. This mixture is passed through a tube containing red hot charcoal. The volume now becomes 1.6 litre. The volumes are measured under the same conditions. Find the composition of the mixture by volume.

(b) A compound contains 28 percent of nitrogen and 72 percent of metal by weight. 3 atoms of metal combine with 2 atoms of N. Find the atomic weight of metal.

1980 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1980

(a) One litre of a sample of hard water contains 1 mg of CaCl₂ and 1 mg of MgCl₂. Find the total hardness in terms of parts of CaCO₃ per 10⁶ parts of water by weight.

(b) A sample of hard water contains 20 mg of Ca⁺⁺ ions per litre. How many milli-equivalent of Na₂CO₃ would be required to soften 1 litre of the sample?

(c) 1 gm of Mg is burnt in a closed vessel which contains 0.5gm of O₂.
      (i) Which reactant is left in excess?
      (ii) Find the weight of the excess reactants.
      (iii) How many milliliters of 0.5N H₂SO₄ will dissolve the residue in the vessel?

1980 JEE Advanced Numerical

iCON Education HYD, 79930 92826, 73309 72826 IIT-JEE 1980

(i). A sample of MnSO₄.4H₂O is strongly heated in air. The residue is Mn₃O₄.

(ii). The residue is dissolved in 100 ml of 0.1N FeSO₄ containing dilute H₂SO₄.

(iii). The solution reacts completely with 50 ml of KMnO₄ solution.

(iv). 25 ml of the KMnO₄ solution used in step (iii) requires 30 ml of 0.1 N FeSO₄ solution for complete reaction.

Find the amount of MnSO₄.4H₂O present in the sample.

Correct Answer: 1.338 g

Explanation:

To find the amount of $ \text{MnSO}_4\cdot4\text{H}_2\text{O} $ present in the sample, we can follow the steps below, applying stoichiometry and the principle of equivalence in titration. Let's analyze each step to solve this problem:

Step (i): Heating MnSO₄.4H₂O

When $ \text{MnSO}_4\cdot4\text{H}_2\text{O} $ is strongly heated, it decomposes, and the end product mentioned is $ \text{Mn}_3\text{O}_4 $. The chemical equation for the decomposition isn't provided directly, but it's crucial to know that we end up with $ \text{Mn}_3\text{O}_4 $ which indicates the reduction of the sulfate.

Step (ii): Dissolving the residue in FeSO₄ solution

The $ \text{Mn}_3\text{O}_4 $ residue reacts with $ \text{FeSO}_4 $ in the presence of dilute $ \text{H}_2\text{SO}_4 $. Here, $ \text{Fe}^{2+} $ acts as a reducing agent.

Step (iii): Reaction with KMnO₄

The solution from step (ii) completely reacts with $ \text{KMnO}_4 $, using 50 ml of $ \text{KMnO}_4 $ solution.

Step (iv): KMnO₄ solution titration

Here, we learn that 25 ml of the $ \text{KMnO}_4 $ solution require 30 ml of 0.1 N $ \text{FeSO}_4 $ for complete reaction. This gives us a direct relationship to calculate the normality (and thus the molarity) of the $ \text{KMnO}_4 $ solution.

To find the normality of the $ \text{KMnO}_4 $ solution, we use:

$ V_1N_1 = V_2N_2 $

where,

$ V_1 $ is the volume of $ \text{FeSO}_4 $ (30 ml),

$ N_1 $ is the normality of $ \text{FeSO}_4 $ (0.1 N),

$ V_2 $ is the volume of $ \text{KMnO}_4 $ (25 ml),

$ N_2 $ is the normality of $ \text{KMnO}_4 $, which we will calculate.

$ 30 \times 0.1 = 25 \times N_2 $

$ N_2 = \frac{30 \times 0.1}{25} = \frac{3}{25} = 0.12\, \text{N} $

Now, knowing that 50 ml of $ \text{KMnO}_4 $ reacts with the $ \text{Mn}_3\text{O}_4 $ formed from the original $ \text{MnSO}_4\cdot4\text{H}_2\text{O} $, and since 25 ml of $ \text{KMnO}_4 $ is 0.12 N, 50 ml of $ \text{KMnO}_4 $ is equivalent to 60 ml of 0.1 N $ \text{FeSO}_4 $. This equivalence helps us understand the amount of $ \text{Fe}^{2+} $ that would react with the $ \text{Mn}^{2+} $ from the original sample.

Given that 60 ml of 0.1 N $ \text{FeSO}_4 $ is used in total, and knowing that 1 mole of $ \text{MnSO}_4 $ reacts to eventually form $ \text{Mn}_3\text{O}_4 $ and is equivalent to the reaction of $ \text{Fe}^{2+} $ with $ \text{KMnO}_4 $, we can calculate the moles of $ \text{MnSO}_4 $ initially present.

$ \text{Moles of } \text{Fe}^{2+} = V \times N = 60 \times 0.1 \times 10^{-3} \text{ L} \times \text{N} = 6 \times 10^{-3} \text{ moles} $

Given the molar ratio of $ \text{Fe}^{2+} $ to $ \text{MnSO}_4 $ in the reactions (which can vary based on the specific reactions involved but typically could be 1:1 in redox reactions involving manganate and iron(II) ions in acidic solution), we can assume that the moles of $ \text{Fe}^{2+} $ directly give us the moles of $ \text{MnSO}_4 $ initially present because each $ \text{Fe}^{2+} $ ion reduces a $ \text{MnO}_4^- $ ion, and the $ \text{Mn}_3\text{O}_4 $ would be linked back to the $ \text{MnSO}_4 $ based on the stoichiometry not given.

Thus, the amount of $ \text{MnSO}_4\cdot4\text{H}_2\text{O} $ present initially can be approximated if we know its molecular weight:

$ \text{MW of } \text{MnSO}_4\cdot4\text{H}_2\text{O} = 151.00 + (4 \times 18.015) = 151.00 + 72.06 = 223.06 \, \text{g/mol} $

Therefore, the mass of $ \text{MnSO}_4\cdot4\text{H}_2\text{O} $ is:

$ \text{Mass} = 6 \times 10^{-3} \text{ moles} \times 223.06 \, \text{g/mol} = 1.33836 \, \text{g} $

In this way, we find that the sample originally contains approximately $ 1.34 $ g of $ \text{MnSO}_4\cdot4\text{H}_2\text{O} $. This calculation is based on an understanding of the stoichiometry of the reactions involved and the titration data. Some assumptions had to be made due to missing precise chemical equations, especially in the conversion between $ \text{MnSO}_4\cdot4\text{H}_2\text{O} $ to $ \text{Mn}_3\text{O}_4 $, and how it reacts with $ \text{KMnO}_4 $ and $ \text{FeSO}_4 $.

Chemistry Chapters

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